Problem: $\dfrac{ x + 3y }{ 7 } = \dfrac{ 3x - 2z }{ -3 }$ Solve for $x$.
Multiply both sides by the left denominator. $\dfrac{ x + 3y }{ {7} } = \dfrac{ 3x - 2z }{ -3 }$ ${7} \cdot \dfrac{ x + 3y }{ {7} } = {7} \cdot \dfrac{ 3x - 2z }{ -3 }$ $x + 3y = {7} \cdot \dfrac { 3x - 2z }{ -3 }$ Multiply both sides by the right denominator. $x + 3y = 7 \cdot \dfrac{ 3x - 2z }{ -{3} }$ $-{3} \cdot \left( x + 3y \right) = -{3} \cdot 7 \cdot \dfrac{ 3x - 2z }{ -{3} }$ $-{3} \cdot \left( x + 3y \right) = 7 \cdot \left( 3x - 2z \right)$ Distribute both sides $-{3} \cdot \left( x + 3y \right) = {7} \cdot \left( 3x - 2z \right)$ $-{3}x - {9}y = {21}x - {14}z$ Combine $x$ terms on the left. $-{3x} - 9y = {21x} - 14z$ $-{24x} - 9y = -14z$ Move the $y$ term to the right. $-24x - {9y} = -14z$ $-24x = -14z + {9y}$ Isolate $x$ by dividing both sides by its coefficient. $-{24}x = -14z + 9y$ $x = \dfrac{ -14z + 9y }{ -{24} }$ Swap signs so the denominator isn't negative. $x = \dfrac{ {14}z - {9}y }{ {24} }$